3 ms·
No you didn't get it. You missed "Let's consider them as operations. Apply any one of them on any other of them, you get the third one."
by zkmon 11mo ago
No you didn't get it. You missed "Let's consider them as operations. Apply any one of them on any other of them, you get the third one."
- youoy 11mo agoSo is what i wrote a third one? Fourth? Fifth? :)
- zkmon 11mo agoNot sure what you are talking about. What you wrote reduces to just x. What I meant was, if you substitute say, -x for x in -1/x, you get 1/x, which is the third inverse. Same is true for the other two pairs. So, if we call them functions f, g and h, then, f=g(h)=h(g); g=f(h)=h(f); h=f(g)=g(f)